The below proof follows e.g. [3]. A proof based on the finite intersection property is given in [4].

Proof. Let $I$ be an indexing set and $F=\{ V_\alpha \mid \alpha \in I\}$ be an arbitrary open cover for $C$ . Since $X\setminus C$ is open, it follows that $F$ together with $X\setminus C$ is an open cover for $K$ . Thus, $K$ can be covered by a finite number of sets, say, $V_1, \ldots, V_N$ from $F$ together with possibly $X\setminus C$ . Since $C\subset K$ , $V_1, \ldots, V_N$ cover $C$ , and it follows that $C$ is compact.

The following proof uses the finite intersection property.

Proof. Let $I$ be an indexing set and $\{A_{\alpha}\}_{\alpha \in I}$ be a collection of $X$ -closed sets contained in $C$ such that, for any finite $J \subseteq I$ , $\displaystyle \bigcap_{\alpha \in J} A_{\alpha}$ is not empty. Recall that, for every $\alpha \in I$ , $A_{\alpha} \subseteq C\subseteq K$ . Thus, for every $\alpha \in I$ , $A_{\alpha}= K\cap A_{\alpha}$ . Therefore, $\{A_{\alpha}\}_{\alpha \in I}$ are $K$ -closed subsets of $K$ (see this page) such that, for any finite $J \subseteq I$ , $\displaystyle \bigcap_{\alpha \in J} A_{\alpha}$ is not empty. As $K$ is compact, $\displaystyle \bigcap_{\alpha \in I} A_{\alpha}$ is not empty (again, by this result). This proves the claim.

Bibliography

1

J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.

2

S. Lang, Analysis II, Addison-Wesley Publishing Company Inc., 1969.

3

G.J. Jameson, Topology and Normed Spaces, Chapman and Hall, 1974.

4

I.M. Singer, J.A. Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.